My take: this puzzle practically seemed to solve itself while filling in cells with their candidates. When that process was done, the vast majority of the cells were solved or occupied with pairs and triples.

At this point, I guessed that the solution could be had via three or four different techniques. However, I noticed the potential for a technique (which I learned from this forum) which, to the best of my knowledge, I've only used two or three times, so I used that one. That solved one cell which was enough to finish it off.

Later, after some discussion, I'll be interested to find out if the technique I used was the "advanced" twist you mentioned.

I agree with you Marty, the puzzle did seem to solve itself. I found three techniques that would finish it off, two of which are essentially the same and make the same exclusion. I think Keith referred to the third.

Solution 1:
Now, notice the remote naked pair <89>, in the chain labeled "a", "b". Thus, R9C3 is not <89>, it must be <5>.

Edit: I see R1C3 is also <89> and could be labeled "b". It achieves the same reduction.

Solution 2:
If you do not see the above, you may notice the interaction: R9 is the only place for <9> in B9. Thus, R9C3 is not <9>. Then, all the unsolved cells have only two candidates, except R8C3 has three candidates.

This is a BUG pattern; R8C3 must be <8>. (If it is not <8>, it is <19>. Then, all unsolved squares have two candidates, and each candidate occurs twice, and only twice, in each row, column, and block. This pattern does not have a unique solution.)

These two I found by hand. Looking at this with Sudoku Susser, there are a number of techniques (coloring, forks, chains) that essentially make the same reductions as the remote pair.

I used the remote pairs. After reading some of the replies, I erased it and did it again to see what options I could find. The first time I missed the locked "9s" in row 9, but noticed it the second time around.

In addition to the remote pairs, I saw coloring on the "8" would do the job, as well as a forcing chain which could probably be based in any unsolved cell.

I did not notice the BUG pattern, and if I did, I wouldn't have known what to do with it.

Quote:

This is a BUG pattern; R8C3 must be <8>. (If it is not <8>, it is <19>. Then, all unsolved squares have two candidates, and each candidate occurs twice, and only twice, in each row, column, and block. This pattern does not have a unique solution.)

So, when you spot a BUG pattern, do you just have to eyeball the grid to see what would create the non-unique solution or is there some sort of guideline? Also, is a BUG pattern ever required to solve a puzzle, or does one use it because it's fun to spot and use? At first thought, it would seem that a puzzle with all two-candidate cells but one would have other solutions.

I used the BUG+1 pattern to solve it, then looking back at it noticed the colouring and the remote pair (grouped colouring). I thought you were referring to the BUG+1 pattern as your 'twist' at the end.

Marty,

In the BUG+1 pattern all unsolved cells have two candidates except for one cell which has three. One of the values in that cell occurs three times in each row, box and column of which that cell is a member. That is the value that must be assigned to that cell to avoid the BUG pattern.

Some call the following a "fork". It is two strong links that line up at one end. (Almost an X-wing.) In this puzzle, it is uses exactly the same cells as the remote pair. Apply it first for <8>, then for <9>.

In this picture, A and B are the only cells in C2 that contain a particular candidate. Similarly, C and D are the only cells in C5 that have the candidate. You can make eliminations of the candidate in the cells marked "*".

If A is true you can make the eliminations. If A is false, B is true, D is false, and C must be true, and you can make the eliminations. There are various names for this (none of which I made up).

The BUG pattern is where every unsolved cell has two candidates. It is not unique.

The BUG+1 pattern is where one cell has three candidates, all the other unsolved cells have two. You have to avoid the BUG.

Actually, I am not a fan of this. If you do not clear all naked and hidden singles first (or, if you do not check that each candidate occurs only twice in each row, column, or box, which is the same thing), The BUG+1 reduction cannot be applied. I have been burned a couple of times.

I probably saw a piece of game 1, as I have for most of the games. I lack the patience to watch whole games, but I tend to catch the last few minutes and watch highlights, as I want to keep abreast of what's happening. I'd be watching the whole game if there was a home team for me. I watch Buffalo Bills games and Syracuse U. basketball games because they are de facto home teams for Rochester.

But Detroit has certainly shown over the last few years how successful the team concept can be.